Replace all occurrences of a string in a data frame

Equivalent to "find and replace." Don't overthink it.

Try it with one:

library(tidyverse)
df <- data.frame(name = rep(letters[1:3], each = 3), var1 = rep('< 2', 9), var2 = rep('<3', 9))

df %>% 
  mutate(var1 = str_replace(var1, " ", ""))
#>   name var1 var2
#> 1    a   <2   <3
#> 2    a   <2   <3
#> 3    a   <2   <3
#> 4    b   <2   <3
#> 5    b   <2   <3
#> 6    b   <2   <3
#> 7    c   <2   <3
#> 8    c   <2   <3
#> 9    c   <2   <3

Apply to all

df %>% 
  mutate_all(funs(str_replace(., " ", "")))
#>   name var1 var2
#> 1    a   <2   <3
#> 2    a   <2   <3
#> 3    a   <2   <3
#> 4    b   <2   <3
#> 5    b   <2   <3
#> 6    b   <2   <3
#> 7    c   <2   <3
#> 8    c   <2   <3
#> 9    c   <2   <3

If the extra space was produced by uniting columns, think about making str_trim part of your workflow.

Created on 2018-03-11 by the reprex package (v0.2.0).

If you are only looking to replace all occurrences of "< " (with space) with "<" (no space), then you can do an lapply over the data frame, with a gsub for replacement:

> data <- data.frame(lapply(data, function(x) {
+                  gsub("< ", "<", x)
+              }))
> data
  name var1 var2
1    a   <2   <3
2    a   <2   <3
3    a   <2   <3
4    b   <2   <3
5    b   <2   <3
6    b   <2   <3
7    c   <2   <3
8    c   <2   <3
9    c   <2   <3